Need help with probability assessment

[swc]

There are 34 family and friends who want to get together next month. If we assume that all of the people come from different areas where, based on population, there is a 1% probability that any 1 person may have an infectious disease, what is the overall likelihood that when we get together that 1 or more people will actually have the disease? I thought it would be = (1-0.01) to the 34th or = 29%. But I have talked to others and have received answers from 1%, 24%, 26% and my 29% and thus I am confused.
The person who answered 1% explained it this way: Assume each person has 1 red tie and 100 yellow ties and they are to bring them to the gathering. When everyone shows up we will pick a tie. What is the probability of picking at least 1 red tie? Isn't the answer 1% (34 out of 3400)? Now, I don't believe this is the answer to my question, but I can not explain why the logic is incorrect.
Can anyone help??
Thanks so much!

 

[Deleted member 4993]

There are 34 family and friends who want to get together next month. If we assume that all of the people come from different areas where, based on population, there is a 1% probability that any 1 person may have an infectious disease, what is the overall likelihood that when we get together that 1 or more people will actually have the disease? I thought it would be = (1-0.01) to the 34th or = 29%. But I have talked to others and have received answers from 1%, 24%, 26% and my 29% and thus I am confused.
The person who answered 1% explained it this way: Assume each person has 1 red tie and 100 yellow ties and they are to bring them to the gathering. When everyone shows up we will pick a tie. What is the probability of picking at least 1 red tie? Isn't the answer 1% (34 out of 3400)? Now, I don't believe this is the answer to my question, but I can not explain why the logic is incorrect.
Can anyone help?? Thanks so much!

Since the question implies at least one person is infected:

We should calculate the probability of NO ONE is infected among 100

select

1 person and the probability that the person does not have infection = 1- 0.01 = 0.99

(1+1=)2 persons and the probability that 2 persons do not have infection = (1- 0.01)^2 = 0.99^2

3 persons and the probability that 3 persons do not have infection = (1- 0.01)^3 = 0.99^3

continue to select 100 persons. Then

select 100 persons and the probability that 100 persons do not have infection = (1- 0.01)^3 = 0.99^100

continue....

 

[Dr.Peterson]

There are 34 family and friends who want to get together next month. If we assume that all of the people come from different areas where, based on population, there is a 1% probability that any 1 person may have an infectious disease, what is the overall likelihood that when we get together that 1 or more people will actually have the disease? I thought it would be = (1-0.01) to the 34th or = 29%. But I have talked to others and have received answers from 1%, 24%, 26% and my 29% and thus I am confused.
The person who answered 1% explained it this way: Assume each person has 1 red tie and 100 yellow ties and they are to bring them to the gathering. When everyone shows up we will pick a tie. What is the probability of picking at least 1 red tie? Isn't the answer 1% (34 out of 3400)? Now, I don't believe this is the answer to my question, but I can not explain why the logic is incorrect.
Can anyone help??
Thanks so much!

Your reasoning is of course correct, though you meant to say 1 - (1 - 0.01)^34 = 0.289.

Using the tie analogy, there is a big difference between picking one tie each from 34 bins of 100, and one of those is red, vs. picking 34 ties from one big bin of 3400, and one of those is red. The latter includes many outcomes that are not part of the former (e.g. picking 34 yellow ties that all would have been in the first bin before they were combined).

And even so, his calculation is nonsense. 34/3400 is the probability that you'd get a red tie if you picked one tie! The probability of picking at least one red out of 34 picks would be 1 - 3366C34/3400C34 = 0.291.

Take a simpler example. Suppose two people have two ties each, one red. If each picks a tie, then the outcomes are RR, RY, YR, YY, and the probability of at least one red is 3/4. If you pick two ties from one set of ABCD, there are 4C2 = 6 ways to choose two ties (AB, AC, AD, BC, BD, CD), of which 5 include at least one red, for a probability of 5/6. How things are chosen matters. (And your friend would probably have answered 2/4, which is even more wrong.)

 

[Steven G]

1 out of a hundred is 1%

So if there are 1 yellow tie and 100 red ties, then it is not correct to say that 1% of the ties are yellow. If there were 1 yellow tie and 99 red ties, then since 1 out of 100 ties are yellow then 1% of the ties are yellow. In your case that you stated you have 1 yellow tie out of 101 which is not 1%.